To solve this problem, we can use Polya's Problem Solving method, which involves these four steps:

1. Understand the Problem:

You are given 10 coins, and one of them is heavier than the rest. You have a balance scale that allows you to compare the weights of two groups of coins. Your goal is to find the minimum number of weighings needed to identify the heavier coin.

2. Devise a Plan:

This problem is a variant of the coin-weighing problem, and the key to minimizing weighings is to divide the coins systematically to reduce the possible number of heavier coins after each weighing. The balance scale has three possible outcomes: left heavier, right heavier, or balanced.

A good approach is to apply the divide-and-conquer method, which minimizes the number of coins at each step.

3. Carry Out the Plan:

We can break the coins into groups and progressively eliminate possibilities based on the results of each weighing.

Step-by-step analysis:

1. Divide the 10 coins into three groups:

   • 3 coins in the first group
   • 3 coins in the second group
   • 4 coins in the third group (or simply leave one coin aside)

2. First weighing: Weigh the first group of 3 coins against the second group of 3 coins.

   • If they balance, the heavier coin must be in the group of 4 coins (including the one set aside).
   • If one side is heavier, the heavier coin is in that group.

3. Second weighing: After the first weighing, you will have reduced the number of coins to a smaller subset of possibilities.

   • If the balance tipped in the first weighing, you are left with 3 coins. Weigh two of them against each other:
     • If one is heavier, you’ve found the coin.
     • If they balance, the heavier coin is the third one.
   • If the first weighing balanced, you are left with 4 coins. Take 3 of these and weigh them against each other:
     • If one side is heavier, the heavier coin is in that group.
     • If they balance, the heavier coin is the one left out.

4. Third weighing (if needed): In the case where you started with 4 coins (if the first weighing balanced), you may need a third weighing to compare the remaining few possibilities.

4. Look Back and Review:

The method described allows you to find the heavier coin in at most 3 weighings. You first reduce the number of coins, then identify the group with the heavier coin, and finally isolate the coin by comparing smaller groups.

Conclusion:

The minimum number of weighings required to find the heavier coin among 10 is 3.

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Would you like further clarification or additional details?

Here are 5 related problems for you to explore:

1. What if you have 27 coins and 1 is heavier—what is the minimum number of weighings needed?
2. How would the solution change if you didn’t know whether the odd coin was heavier or lighter?
3. What is the minimum number of weighings for 9 coins if one is either heavier or lighter?
4. What happens if you have 12 coins with one odd coin (heavier or lighter)?
5. How does the strategy change if you can only use the balance scale twice?

Tip: When solving coin-weighing problems, think of each weighing as a way to split the possibilities into three outcomes: heavier, lighter, or balanced. This helps narrow down the possibilities faster!